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  Fourier transform of the G-lesser function.

+ 1 like - 0 dislike

I am studying Kadanoff & Baym's book Quantum Statistical Mechanics and I am stuck a one point.

The are considering a system of non-interacting particles, (let's say fermions to not having to write both signs), and are then considering the G-lesser function:

$$ G^{<} (1, 1') =  i \left< \psi^\dagger (1') \psi(1) \right> , $$

where $ 1 = \mathbf{r}_1, t_1 $ and similarly for $1'$.

Since the Hamiltonian has rotational and translational symmetry they argue that the Green's function above only depends on $| \mathbf{r}_1 - \mathbf{r}_{1'} |$. Also, since the Hamiltonian is time independent the Green's function should only depend on the time difference $t_1 - t_1'$. All this seems fine and I think I have sucessfully convinced myself of these facts by considering e.g. the translation operator.

However, they then define the Fourier transform as

$$ G^{<} ( \mathbf{p}, \omega) = - i \int d^3 r \int dt e^{-i \mathbf{p} \cdot \mathbf{r} + i \omega t} G^{<}(\mathbf{r}, t),  $$

where we now use $ \mathbf r = \mathbf r_1 - \mathbf r_2$ and similarly for $t$. Now come the claim that I cannot really see. They say that, due to the invariances I talked about above, we have

$$ G^{<}(\mathbf{p} , \omega) =
\int dt \frac{e^{i\omega t}}{V} \left< \psi^\dagger(\mathbf{p}, 0) \psi(\mathbf{p}, t) \right> $$

where $V$ is the volume of the system. Can someone please explain how this follows from the above? If I naively try to calculate this I instead get

$$ G^{<} ( \mathbf{p}, \omega) = - i \int d^3 r \int dt e^{-i \mathbf{p} \cdot \mathbf{r} + i \omega t} G^{<}(\mathbf{r}, t)
\\ = - i \int d^3 r \int dt e^{-i \mathbf{p} \cdot \mathbf{r} + i \omega t} \left< \psi^\dagger(\mathbf{0}, 0) \psi(\mathbf{r}, t ) \right>, $$

which would only give the Fourier transform of the annihilation operator.

I have tried to redo everything carefully from scratch and I have found for example the relation

$$ G^{<} ( \mathbf{p}, \omega, \mathbf{p}', \omega' ) = 2\pi V \delta(\omega - \omega') \delta_{\mathbf{p}, \mathbf{p}'} G^{<} ( \mathbf{p}, \omega). $$

However, I cannot really get the factor $1/V$ which was obtained in the book.

asked Feb 12 in Theoretical Physics by JezuzStardust (5 points) [ revision history ]

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